Rainbows go in the order of ROYGBIV, which is an acronym for red, orange, yellow, green, blue, indigo, violet.
So, it would start off with red.
If you want to know why it starts off the rainbow, its because red has the longest wavelength, compared to the rest of the colors. (sorry I'm rambling, I got excited)
Hope this helps.
Answer : When we consider the atmospheric pressure as 1 atm then according to the ideal gas equation we can find out the molar mass of any unknown by this formula ;
PV=nRT
so if the pressure increases than 1 atm then we can see from the above equation that it will result in greater value for the number of moles (n) in the above equation.
While n = m/M where m is mass of the unknown in g and M is molecular mass.
So, if pressure is higher then it will result in molar mass of unknown which is much smaller.
I found the attached image with the same statement of your question and think it may be very useful for you that I use it to show how to answer this question (furthermore I think it may be the same reaction that you forgot to include).
As you can see, there are one image on the left side and other image on hte right side of the figure.
Those images contains drawings that represent molecules and a legend that permit you to distinguish the kind of atoms in each molecule.
Using that, you can indicate the chemical reaction as the transformation of the molecules on the left side onto the molecules on the right side:
Left side:
3 molecules of CH4 and 3 molecules of N2Cl4
Right side:
3 molecules of CCl4, 3 molecules of N2 and 6 molecules of H2
That is represented as:
3CH4 + 3 N2Cl4 -----> 3 CCl4 + 3N2 + 6H2And that is the balanced chemical equation for the reaction shown in the figured attached.I hope this is useful for you..
The question is incomplete, the complete question is;
Why is a terminal alkyne favored when sodium amide (NaNH2) is used in an elimination reaction with 2,3-dichlorohexane? product. A) The terminal alkyne is more stable than the internal alkyne and is naturally the favored B) The terminal alkyne is not favored in this reaction. C) The resonance favors the formation of the terminal rather than internal alkyne. D) The strong base deprotonates the terminal alkyne and removes it from the equilibrium.
E) The positions of the Cl atoms induce the net formation of the terminal alkyne.
Answer:
E) The positions of the Cl atoms induce the net formation of the terminal alkyne.
Explanation:
In this reaction, sterric hindrance plays a very important role. We know that sodamide is a strong base, it tends to attack at the most accessible position.
The first deprotonation yields an alkene. The strong base attacks at the terminal position again and yields the terminal alkyne. Thus the structure of the dihalide makes the terminal hydrogen atoms most accessible to the base. Hence the answer.
Answer:
Explanation:
a. What is the mass number of the particle emitted from the nucleus during beta minus (β–) decay?
zero
The beta radiations are emitted in this reaction. The one electron is ejected and neutron is converted into proton.
⁴₆C → ¹⁴₇N + ⁰₋₁e
b. What kind of charge does the particle emitted from the nucleus during beta minus (β–) decay have?
Negative charge
Electron is emitted during beta decay and it carry negative charge.
c. What is another name for a beta minus (β–) particle?
Electron
During beta minus decay electron is emitted and neutron is converted into proton.
d. Write the balanced equation for the alpha decay that is below the “Show Equation.” Label the parent, daughter, and beta particle.
Equation is missing
a. What happens in the nucleus of an atom when an alpha particle is emitted?
When atom undergoes the alpha emission the original atom convert into the atom having mass number less than 4 and atomic number less than 2 as compared to the starting atom.
b. b. What happens in the nucleus of an atom when a beta particle is emitted?
When nucleus emit the beta particle neutron is converted into proton and this proton stay into the nucleus while at the same time electron is emitted. Thus atomic number is increased by one.
⁴₆C → ¹⁴₇N + ⁰₋₁e
